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My Algebra book says that the category Fld of fields has no initial object.

Why would $\{0,1\}$ not be an initial object? Does it not have a unique homomorphism to every other field?

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How would it map into $\mathbb Z_3$? and if you have an answer to that, then why won't it map in two different ways into $\mathbb Z_2$? – Ittay Weiss Jun 27 '14 at 2:35

Any field fails to embed in fields of different characteristic. (Consider this a basic exercise.)

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Oh crap, 30 seconds! :-) – Asaf Karagila Jun 27 '14 at 2:36
@AsafKaragila: boy, do I know that feeling! ;-) – Robert Lewis Jun 27 '14 at 3:43
@Robert: I'm usually on the faster side. Having a good typing rate, and a relatively low typo rate as well. Strange to be on the other side for once! ;-) – Asaf Karagila Jun 27 '14 at 3:52
@AsafKaragila: I am one of the fastest typists most people I know have ever seen, and I still "come in second" now and then. When I worked in telecomm my first day on the job one of my colleagues in engineering said I was the fastest typist he'd ever seen! Must be connected to finger-work on the guitar! ;-) I usually don't worry too much about being first, since most of my answers have been unique enough to post no matter when they get done. Anyway, I've given up "racing" by and large; except, of course, on unpoliced highways! (Continued . . .) – Robert Lewis Jun 27 '14 at 4:12
@AsafKaragila: (Continuation) What really bugs me is question closure just when I have great answer I'm about to post! But what the hey, we're all sporting gents here, no? Cheers! – Robert Lewis Jun 27 '14 at 4:15

The obvious mapping would seem to be $0\mapsto0$ and $1\mapsto1$, but that implies $1+1\mapsto0$, and so $1+1$ would have no multiplicative inverse. A homomorphism would map inverses to inverses.

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It is perhaps worth noting that, in some texts, the definition of a field does not require the additive and multiplicative identities to be distinct. It isn't difficult to verify that $\{0\}$ satisfies the field axioms if we remove this restriction, and in particular, such texts would consider $\{0\}$ to be precisely the initial object of $\mathbf{Fld}.$

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I have long claimed that someone needs to take all the known mathematics and do it over the field of one element, $\{0\}$. After all, it is the unique ordered field which is both finite and complete! :-) – Asaf Karagila Jun 28 '14 at 16:46
It's the only field of characteristic 1, too. – Cameron Buie Jun 28 '14 at 17:23

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